In the redox reaction, taking place in acidic medium: X MnO ₄^- (aq) + Y SO ₂(g) Mn ⁺²(aq) + HSO ₄^- (aq) , the ratio of X:Y in a stoichiometrically balanced equation will be

Step-by-step solution

The given redox reaction can be balanced by splitting it into two half-reactions. Reduction half-reaction: MnO ₄^- + 8 H ^+ + 5e^- Mn ⁺² + 4 H ₂ O Oxidation half-reaction: SO ₂ + 2 H ₂ O HSO ₄^- + 3 H ^+ + 2e^- To equalize the number of electrons transferred, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 : 2 MnO ₄^- + 16 H ^+ + 10e^- 2 Mn ⁺² + 8 H ₂ O 5 SO ₂ + 10 H ₂ O 5 HSO ₄^- + 15 H ^+ + 10e^- Adding the two balanced half-reactions, we get the overall balanced equation: 2 MnO ₄^-